Hyroko
  • Hyroko
How do I do implicit differentiation on the equation x^2 -4xy +y^2 = 4 ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
welshfella
  • welshfella
treat y as a function of x so the derivative with respect to x, of y^2 is 2y.dy/dx use the product rule to differentiate 4xy find the derivative term by term then make dy/dx the subject
welshfella
  • welshfella
does this help?
Hyroko
  • Hyroko
I get Y -4dy/dx for the -4xy. is this correct?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

welshfella
  • welshfella
no its -(4x * dy/dx + 4y) = -4x dy/dx - 4y
welshfella
  • welshfella
so we have 2x - 4x.dy/dx - 4y + 2y.dy/dx = 0 now solve for dy/dx
Hyroko
  • Hyroko
I get Y Prime = (x+3y)/ -2x I know this is not the answer
Hyroko
  • Hyroko
I should be getting (2y-x)/(y-2x)
Hyroko
  • Hyroko
But I'm not.
Hyroko
  • Hyroko
because I know the answer, but that does me no good if I don't understand how to get the answer. The part that is tripping me up the most is that -4xy. we use y' instead of dy/dx in our class.
welshfella
  • welshfella
2x - 4x.dy/dx - 4y + 2y.dy/dx = 0 2y.dy/dx - 4x dydx = 4y - 2x dy/dx = (4y - 2x) / ( 2y - 4x) dividing top and bottm by 2 dy/dx = (2y - x)/(y - 2x)
anonymous
  • anonymous
can you explain how you get to your answer
welshfella
  • welshfella
derivative of -4xy = derivative of - (4xy) = - ( 4x.y' + y* 4) = -4xy' - 4y
welshfella
  • welshfella
Product Rule: f'(xy) = f(x)f'(y) + f(y)f'(x)
welshfella
  • welshfella
the derivative of y^2 is found using the chain rule
Hyroko
  • Hyroko
it's mainly the algebra that is messing me up . being able to get 2x - 4xy'-4y=2yy' into the form (2y-x)/ (y-2x)
Hyroko
  • Hyroko
I am having trouble getting the Y' by itself on one side.
welshfella
  • welshfella
2y.dy/dx - 4x dydx = 4y - 2x = 2y y' - 4x y' = 4y - 2x factor the left side (y' is common to both terms):- y' (2y - 4x) = 4y - 2x Now divide both sides by 2y-4x:- y' = (4y - 2x)/(2y - 4x) dividing by 2 gives y' = (2y - x)/(y - 2x)
welshfella
  • welshfella
* dividing top an d bottom of fraction by 2
Hyroko
  • Hyroko
alright. that clears it up. thank you
welshfella
  • welshfella
yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.